Problem: Simplify the following expression: $y = \dfrac{7x^2- 29x+4}{7x - 1}$
Solution: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(7)}{(4)} &=& 28 \\ {a} + {b} &=& &=& {-29} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $28$ and add them together. The factors that add up to ${-29}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-1}$ and ${b}$ is ${-28}$ $ \begin{eqnarray} {ab} &=& ({-1})({-28}) &=& 28 \\ {a} + {b} &=& {-1} + {-28} &=& -29 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({7}x^2 {-1}x) + ({-28}x +{4}) $ Factor out the common factors: $ x(7x - 1) - 4(7x - 1)$ Now factor out $(7x - 1)$ $ (7x - 1)(x - 4)$ The original expression can therefore be written: $ \dfrac{(7x - 1)(x - 4)}{7x - 1}$ We are dividing by $7x - 1$ , so $7x - 1 \neq 0$ Therefore, $x \neq \frac{1}{7}$ This leaves us with $x - 4; x \neq \frac{1}{7}$.